AMC12 2004 B

AMC12 2004 B · Q4

AMC12 2004 B · Q4. It mainly tests Basic counting (rules of product/sum), Inclusion–exclusion (basic).

An integer $x$, with $10\leq x\leq 99$, is to be chosen. If all choices are equally likely, what is the probability that at least one digit of $x$ is a 7?

要选择一个整数 $x$，满足 $10\leq x\leq 99$。如果所有选择等可能，至少有一位数字是 7 的概率是多少？

(A) \frac{1}{9} \frac{1}{9}

(B) \frac{1}{5} \frac{1}{5}

(D) \frac{2}{9} \frac{2}{9}

(E) \frac{1}{3} \frac{1}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

The digit $7$ can be either the tens digit ($70, 71, \dots, 79$: $10$ possibilities), or the ones digit ($17, 27, \dots, 97$: $9$ possibilities), but we counted the number $77$ twice. This means that out of the $90$ two-digit numbers, $10+9-1=18$ have at least one digit equal to $7$. Therefore the probability is $\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}$.

数字 $7$ 可以出现在十位（$70, 71, \dots, 79$：$10$ 种可能），也可以出现在个位（$17, 27, \dots, 97$：$9$ 种可能），但数 $77$ 被重复计算了一次。这意味着在 $90$ 个两位数中，有 $10+9-1=18$ 个至少有一位数字等于 $7$。因此概率为 $\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}$。

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